Question: A curve in the plane is defined parametrically by the equations $x=\ln(1-2t)$ and $y=\dfrac{1}{3t}$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{2}{\ln(1-2t)}$ (Choice B) B $-\dfrac{1}{3t^2}$ (Choice C) C $\dfrac{1-2t}{9t^2}$ (Choice D) D $\dfrac{1-2t}{6t^2}$
In general, to find the derivative (i.e. the expression for $\dfrac{dy}{dx}$ ) of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ (where $u$ and $v$ are any functions of $t$ ), we use the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ We are given that $x=\ln(1-2t)$ and $y=\dfrac{1}{3t}$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\dfrac{d}{dt}\left(\dfrac{1}{3t}\right)}{\dfrac{d}{dt}(\ln(1-2t))} \\\\ &=\dfrac{-\dfrac{1}{3t^2}}{-\dfrac{2}{1-2t}} \\\\ &=\dfrac{1-2t}{6t^2} \gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{1-2t}{6t^2}$.